thermosolutions chapter12

12-1

Chapter 12 THERMODYNAMIC PROPERTY RELATIONS

Partial Derivatives and Associated Relations

12-1C

yx zzdzdyydxx

12-2C For functions that depend on one variable, they are identical. For functions that depend on two or more variable, the partial differential represents the change in the function with one of the variables as theother variables are held constant. The ordinary differential for such functions represents the total change as a result of differential changes in all variables.

12-3C (a) ( x)y = dx ; (b) ( z) y dz; and (c) dz = ( z)x + ( z) y

12-4C Only when ( z/ x) y = 0. That is, when z does not depend on y and thus z = z(x).

12-5C It indicates that z does not depend on y. That is, z = z(x).

12-6C Yes.

12-7C Yes.

y

( z)x

( z)y

dx

dz

x +dx

y + dy

z

dyx

y

x

12-2

12-8 Air at a specified temperature and specific volume is considered. The changes in pressurecorresponding to a certain increase of different properties are to be determined.Assumptions Air is an ideal gasProperties The gas constant of air is R = 0.287 kPa·m3/kg·K (Table A-1). Analysis An ideal gas equation can be expressed as P = RT/v. Noting that R is a constant and P = P(T,v),

2v

v

vvv

dTRdTRdv

T

PdTTPdP

(a) The change in T can be expressed as dT T = 400 0.01 = 4.0 K. At v = constant,

kPa276.1/kgm0.90

K)K)(4.0/kgmkPa(0.2873

3

vvdTR

dP

(b) The change in v can be expressed as dv v = 0.90 0.01 = 0.009 m3/kg. At T = constant,

kPa276.123

33

2 /kg)m(0.90/kg)mK)(0.009K)(400/kgmkPa(0.287

v

vdTRdP T

(c) When both v and T increases by 1%, the change in P becomes0)276.1(276.1)()( TdPdPdP v

Thus the changes in T and v balance each other.

12-9 Helium at a specified temperature and specific volume is considered. The changes in pressure corresponding to a certain increase of different properties are to be determined.Assumptions Helium is an ideal gasProperties The gas constant of helium is R = 2.0769 kPa·m3/kg·K (Table A-1). Analysis An ideal gas equation can be expressed as P = RT/v. Noting that R is a constant and P = P(T,v ),

2v

v

vv

vv

dTRdTRd

T

PdTTPdP

(a) The change in T can be expressed as dT T = 400 0.01 = 4.0 K. At v = constant,

kPa2319./kgm0.90

K)K)(4.0/kgmkPa(2.07693

3

vvdTR

dP

(b) The change in v can be expressed as dv v = 0.90 0.01 = 0.009 m3/kg. At T = constant,

kPa2319./kg)m(0.90

)mK)(0.009K)(400/kgmkPa(2.076923

33

2v

vdTRdP T

(c) When both v and T increases by 1%, the change in P becomes0)231.9(231.9)()( TdPdPdP v

Thus the changes in T and v balance each other.

12-3

12-10 It is to be proven for an ideal gas that the P = constant lines on a T-v diagram are straight lines and that the high pressure lines are steeper than the low-pressure lines. Analysis (a) For an ideal gas Pv = RT or T = Pv/R. Taking the partial derivative of T with respect to vholding P constant yields

RPT

PvT

P = const

v

which remains constant at P = constant. Thus the derivative( T/ v)P, which represents the slope of the P = const. lines on a T-v diagram, remains constant. That is, the P = const. lines are straight lines on a T-v diagram.(b) The slope of the P = const. lines on a T-v diagram is equal toP/R, which is proportional to P. Therefore, the high pressure linesare steeper than low pressure lines on the T-v diagram.

12-11 A relation is to be derived for the slope of the v = constant lines on a T-P diagram for a gas that obeys the van der Waals equation of state.Analysis The van der Waals equation of state can be expressed as

baPR

T vv 2

1

Taking the derivative of T with respect to P holding v constant,

Rbb

RPT v

vv

011

which is the slope of the v = constant lines on a T-P diagram.

12-4

12-12 Nitrogen gas at a specified state is considered. The cp and cv of the nitrogen are to be determinedusing Table A-18, and to be compared to the values listed in Table A-2b.Analysis The cp and cv of ideal gases depends on temperature only, and are expressed as cp(T) = dh(T)/dTand cv(T) = du(T)/dT. Approximating the differentials as differences about 400 K, the cp and cv values aredetermined to be

KkJ/kg1.045

K390)(410kJ/kg.011,347)/28(11,932

K390410K390K410

)()()K400(K400K400

hh

TTh

dTTdhc

TTp cp

h

T(Compare: Table A-2b at 400 K cp = 1.044 kJ/kg·K)

KkJ/kg0.748K390)(410

kJ/kg08,104)/28.(8,523

K390410K390K410

)()()K400(K400K400

uu

TTu

dTTduc

TTv

(Compare: Table A-2b at 400 K cv = 0.747 kJ/kg·K)

12-13E Nitrogen gas at a specified state is considered. The cp and cv of the nitrogen are to be determinedusing Table A-18E, and to be compared to the values listed in Table A-2Eb.Analysis The cp and cv of ideal gases depends on temperature only, and are expressed as cp(T) = dh(T)/dTand cv(T) = du(T)/dT. Approximating the differentials as differences about 600 R, the cp and cv values are determined to be

RBtu/lbm0.249R580)(620

Btu/lbm8.04,028.7)/2(4,307.1

R580620R580R620

)()()R600(R600R600

hh

TTh

dTTdhc

TTp

(Compare: Table A-2Eb at 600 R cp = 0.248 Btu/lbm·R )

RBtu/lbm0.178R580)(620

Btu/lbm8.02,876.9)/2(3,075.9

R580620R580R620

)()()R600(R600R600

uu

TTu

dTTduc

TTv

(Compare: Table A-2Eb at 600 R cv = 0.178 Btu/lbm·R)

12-5

12-14 The state of an ideal gas is altered slightly. The change in the specific volume of the gas is to be determined using differential relations and the ideal-gas relation at each state. Assumptions The gas is air and air is an ideal gas. Properties The gas constant of air is R = 0.287 kPa·m3/kg·K (Table A-1). Analysis (a) The changes in T and P can be expressed as

kPa4kPa100)96(K4K400)(404

PdPTdT

The ideal gas relation Pv = RT can be expressed as v = RT/P. Note that R is a constant and v = v (T, P).Applying the total differential relation and using average values for T and P,

/kgm0.0598 3/kg)m(0.04805/kg)m(0.0117

kPa)(98kPa)4K)((402

kPa98K4

K)/kgmkPa(0.287

33

23

2PdPTR

PdTR

dPP

dTT

dTP

vvv

(b) Using the ideal gas relation at each state,

/kgm1.2078kPa96

K)K)(404/kgmkPa(0.287

/kgm1.1480kPa100

K)K)(400/kgmkPa(0.287

33

2

22

33

1

11

PRT

PRT

v

v

Thus,

/kgm0.0598 31480.12078.112 vvv

The two results are identical.

12-6

12-15 Using the equation of state P(v-a) = RT, the cyclic relation, and the reciprocity relation at constant vare to be verified. Analysis (a) This equation of state involves three variables P, v, and T. Any two of these can be taken as the independent variables, with the remaining one being the dependent variable. Replacing x, y, and z byP, v, and T, the cyclic relation can be expressed as

1v

vv P

TT

P

PT

where

Ra

PT

RaPT

PR

Ta

PRT

aP

aRTP

aRTP

P

T

vv

vv

vvvv

v

)(

2

Substituting,

1R

aPR

aP

PT

TP

PT

vv

vv v

which is the desired result.(b) The reciprocity rule for this gas at v = constant can be expressed as

aR

TP

aRTP

Ra

PT

RaPT

PTTP

vv

vv

v

v

vv

)()/(

1

We observe that the first differential is the inverse of the second one. Thus the proof is complete.

12-7

The Maxwell Relations

12-16 The validity of the last Maxwell relation for refrigerant-134a at a specified state is to be verified.Analysis We do not have exact analytical property relations for refrigerant-134a, and thus we need toreplace the differential quantities in the last Maxwell relation with the corresponding finite quantities. Using property values from the tables about the specified state,

K/kgm101.0095K/kgm101.005

C60)(100/kgm0.018404)(0.022442

kPa1000)(1400KkJ/kg1.0458)(1.0056

C60100kPa10001400

3434

3?

kPa1200

C60C100?

C80

kPa1000kPa1400

kPa1200

?

C80

?

PT

PT

PT

ss

TPs

TPs

vv

v

v

since kJ kPa·m³, and K C for temperature differences. Thus the last Maxwell relation is satisfied.

12-8

12-17 EES Problem 12-16 is reconsidered. The validity of the last Maxwell relation for refrigerant 134a atthe specified state is to be verified. Analysis The problem is solved using EES, and the solution is given below.

"Input Data:"

T=80 [C] P=1200 [kPa] P_increment = 200 [kPa] T_increment = 20 [C] P[2]=P+P_incrementP[1]=P-P_incrementT[2]=T+T_incrementT[1]=T-T_increment

DELTAP = P[2]-P[1] DELTAT = T[2]-T[1]

v[1]=volume(R134a,T=T[1],P=P)v[2]=volume(R134a,T=T[2],P=P)s[1]=entropy(R134a,T=T,P=P[1])s[2]=entropy(R134a,T=T,P=P[2])

DELTAs=s[2] - s[1]DELTAv=v[2] - v[1]

"The partial derivatives in the last Maxwell relation (Eq. 11-19) is associated with the Gibbsfunction and are approximated by the ratio of ordinary differentials:"

LeftSide =DELTAs/DELTAP*Convert(kJ,m^3-kPa) "[m^3/kg-K]" "at T = Const."RightSide=-DELTAv/DELTAT "[m^3/kg-K]" "at P = Const."

SOLUTION

DELTAP=400 [kPa] DELTAs=-0.04026 [kJ/kg-K] DELTAT=40 [C] DELTAv=0.004038 [m^3/kg] LeftSide=-0.0001007 [m^3/kg-K] P=1200 [kPa] P[1]=1000 [kPa] P[2]=1400 [kPa] P_increment=200 [kPa]

RightSide=-0.000101 [m^3/kg-K] s[1]=1.046 [kJ/kg-K] s[2]=1.006 [kJ/kg-K] T=80 [C] T[1]=60 [C] T[2]=100 [C] T_increment=20 [C] v[1]=0.0184 [m^3/kg] v[2]=0.02244 [m^3/kg]

12-9

12-18E The validity of the last Maxwell relation for steam at a specified state is to be verified. Analysis We do not have exact analytical property relations for steam, and thus we need to replace the differential quantities in the last Maxwell relation with the corresponding finite quantities. Using property values from the tables about the specified state,

R/lbmft101.635R/lbmft101.639

F00)7(900/lbmft.6507)1(1.9777

psia0)35(450RBtu/lbm1.7009)(1.6706

F007900psia035450

3333

3?

psia400

F700F900?

F800

psia350psia450

psia400

?

F800

?

PT

PT

PT

ss

TPs

TPs

vv

v

v

since 1 Btu 5.4039 psia·ft3, and R F for temperature differences. Thus the fourth Maxwell relation issatisfied.

12-19 Using the Maxwell relations, a relation for ( s/ P)T for a gas whose equation of state is P(v-b) = RTis to be obtained.

Analysis This equation of state can be expressed as bP

RTv . Then,

PR

T P

v

From the fourth Maxwell relation,

PR

PT TPs v

12-20 Using the Maxwell relations, a relation for ( s/ v)T for a gas whose equation of state is (P-a/v2)(v-b)= RT is to be obtained.

Analysis This equation of state can be expressed as 2vv

ab

RTP . Then,

bR

TP

vv

From the third Maxwell relation,

bR

vv vTPs

T

12-10

12-21 Using the Maxwell relations and the ideal-gas equation of state, a relation for ( s/ v)T for an ideal gas is to be obtained.

Analysis The ideal gas equation of state can be expressed as v

RTP . Then,

vv

RTP

From the third Maxwell relation,

vv v

RTPs

T

The Clapeyron Equation

12-22C It enables us to determine the enthalpy of vaporization from hfg at a given temperature from the P,v, T data alone.

12-23C It is exact.

12-24C It is assumed that vfg vg RT/P, and hfg constant for small temperature intervals.

12-25 Using the Clapeyron equation, the enthalpy of vaporization of refrigerant-134a at a specifiedtemperature is to be estimated and to be compared to the tabulated data. Analysis From the Clapeyron equation,

kJ/kg163.00K4

kPa.68)963(1072.8/kg)m0.00087202K)(0.01995273.15(40

C83C42)(

)(

3

C83@satC42@satC04@

C04sat,C04@

sat

PPT

TPT

dTdPTh

fg

fg

fgfg

vv

vv

v

The tabulated value of hfg at 40°C is 163.00 kJ/kg.

12-11

12-26 EES Problem 12-25 is reconsidered. The enthalpy of vaporization of refrigerant 134-a as a functionof temperature over the temperature range -20 to 80°C by using the Clapeyron equation and the refrigerant134-a data in EES is to be plotted.Analysis The problem is solved using EES, and the results are tabulated and plotted below.

"Input Data:"

T=30 [C] T_increment = 5 [C]

-20 0 20 40 60 800.05

0.1

0.15

0.2

0.25

0.3

0.35

T [C]

PercentError[%]

T[2]=T+T_incrementT[1]=T-T_incrementP[1] = pressure(R134a,T=T[1],x=0) P[2] = pressure(R134a,T=T[2],x=0) DELTAP = P[2]-P[1] DELTAT = T[2]-T[1]

v_f=volume(R134a,T=T,x=0)v_g=volume(R134a,T=T,x=1)h_f=enthalpy(R134a,T=T,x=0)h_g=enthalpy(R134a,T=T,x=1)

h_fg=h_g - h_f v_fg=v_g - v_f

"The Clapeyron equation (Eq. 12-22) provides a means to calculate the enthalpy of vaporization,h_fg at a given temperature by determining the slope of the saturation curve on a P-T diagramand the specific volume of the saturated liquid and satruated vapor at the temperature."

h_fg_Clapeyron=(T+273.2)*v_fg*DELTAP/DELTAT*Convert(m^3-kPa,kJ)PercentError=ABS(h_fg_Clapeyron-h_fg)/h_fg*Convert(, %) "[%]"

hfg

[kJ/kg]hfg,Clapeyron

[kJ/kg]PercentError

[%] T

[C]212.91 213.68 0.3593 -20205.96 206.56 0.2895 -10198.60 199.05 0.2283 0190.73 191.07 0.1776 10182.27 182.52 0.1394 20173.08 173.28 0.1154 30163.00 163.18 0.1057 40151.79 151.96 0.1081 50139.10 139.26 0.1166 60124.32 124.47 0.1195 70106.35 106.45 0.09821 80

12-12

12-27 Using the Clapeyron equation, the enthalpy of vaporization of steam at a specified pressure is to beestimated and to be compared to the tabulated data.Analysis From the Clapeyron equation,

kJ/kg2161.1C130.58)(136.27

kPa50/kg)m0.001073K)(0.60582273.15(133.52

kPa75)2(325)(

)(

3

kPa@275satkPa253@satkPa003@kPa300@sat

kPa300sat,kPa300@

sat

TTT

TPT

dTdPTh

fg

fg

fgfg

vv

vv

v

The tabulated value of hfg at 300 kPa is 2163.5 kJ/kg.

12-28 The hfg and sfg of steam at a specified temperature are to be calculated using the Clapeyron equationand to be compared to the tabulated data. Analysis From the Clapeyron equation,

kgkJ82206 /.K10

kPa.18)169(232.23/kg)m0.001060K)(0.89133273.15(120

C115C125)(

)(

3

C115@satC125@satC120@

C120sat,C120@

sat

PPT

TPT

dTdPTh

fg

fg

fgfg

vv

vv

v

Also,

Kkg/kJ6131.5K273.15)(120

kJ/kg2206.8T

hs fg

fg

The tabulated values at 120 C are hfg = 2202.1 kJ/kg and sfg = 5.6013 kJ/kg·K.

12-13

12-29E [Also solved by EES on enclosed CD] The hfg of refrigerant-134a at a specified temperature is to be calculated using the Clapeyron equation and Clapeyron-Clausius equation and to be compared to thetabulated data.Analysis (a) From the Clapeyron equation,

error)(0.2%/lbmftpsia444.0

R20psia49.776)(72.152

/lbm)ft0.01270R)(0.79136459.67(50

F40F60)(

)(

3

3

F40@satF60@satF50@

F50sat,F50@

sat

Btu/lbm82.16

PPT

TPT

dTdPTh

fg

fg

fgfg

vv

vv

v

since 1 Btu = 5.4039 psia·ft3.(b) From the Clapeyron-Clausius equation,

error)(14.4%

R67.459601

R67.459401

RBtu/lbm0.01946psia776.49psia72.152

ln

11lnsat21sat1

2

Btu/lbm93.80fg

fg

fg

h

h

TTRh

PP

The tabulated value of hfg at 50 F is 82.00 Btu/lbm.

12-14

12-30 EES The enthalpy of vaporization of steam as a function of temperature using Clapeyron equationand steam data in EES is to be plotted.Analysis The enthalpy of vaporization is determined using Clapeyron equation from

TPTh fgfg vClapeyron,

At 100ºC, for an increment of 5ºC, we obtain

/kgm6710.1001043.06720.1

/kgm6720.1

/kgm001043.0kPa29.3661.8490.120

C1095105kPa90.120

kPa61.84C1055100

C955100

3

3C100@

3C100@

12

12

C105@sat2

C95@sat1

increment2

increment1

fgfg

g

f

PPPTTT

PPPP

TTTTTT

vvv

v

v

Substituting,

kJ/kg2262.8K10kPa36.29/kg)mK)(1.671015.273100( 3

Clapeyron, TPTh fgfg v

The enthalpy of vaporization from steam table is

/kgm2256.4 3C100@fgh

The percent error in using Clapeyron equation is

0.28%1004.2256

4.22568.2262orPercentErr

We repeat the analysis over the temperature range 10 to 200ºC using EES. Below, the copy of EES solution is provided:

"Input Data:""T=100" "[C]"T_increment = 5"[C]"T[2]=T+T_increment"[C]"T[1]=T-T_increment"[C]"P[1] = pressure(Steam_iapws,T=T[1],x=0)"[kPa]"P[2] = pressure(Steam_iapws,T=T[2],x=0)"[kPa]"DELTAP = P[2]-P[1]"[kPa]"DELTAT = T[2]-T[1]"[C]"

v_f=volume(Steam_iapws,T=T,x=0)"[m^3/kg]"v_g=volume(Steam_iapws,T=T,x=1)"[m^3/kg]"h_f=enthalpy(Steam_iapws,T=T,x=0)"[kJ/kg]"h_g=enthalpy(Steam_iapws,T=T,x=1)"[kJ/kg]"

h_fg=h_g - h_f"[kJ/kg-K]"v_fg=v_g - v_f"[m^3/kg]"

12-15

"The Clapeyron equation (Eq. 11-22) provides a means to calculate the enthalpy of vaporization,h_fg at a given temperature by determining the slope of the saturation curve on a P-T diagramand the specific volume of the saturated liquid and satruated vapor at the temperature."

h_fg_Clapeyron=(T+273.15)*v_fg*DELTAP/DELTAT*Convert(m^3-kPa,kJ)"[kJ/kg]"PercentError=ABS(h_fg_Clapeyron-h_fg)/h_fg*100"[%]"

hfg

[kJ/kg]hfg,Clapeyron

[kJ/kg]PercentErro

r[%]

T[C]

2477.20 2508.09 1.247 102429.82 2451.09 0.8756 302381.95 2396.69 0.6188 502333.04 2343.47 0.4469 702282.51 2290.07 0.3311 902229.68 2235.25 0.25 1102173.73 2177.86 0.1903 1302113.77 2116.84 0.1454 1502014.17 2016.15 0.09829 1801899.67 1900.98 0.06915 2101765.50 1766.38 0.05015 240

0 50 100 150 200 2501700

1800

1900

2000

2100

2200

2300

2400

2500

2600

T [C]

hfg

[kJ/kg]

hfg calculated by EES

hfg calculated by Clapeyron equation

12-16

12-31 The sublimation pressure of water at -30ºC is to be determined using Clapeyron-Clasius equationand the triple point data of water.Analysis The sublimation pressure may be determined using Clapeyron-Clasius equation from

211

CCsub, 11lnTTR

hP

P ig

where the triple point properties of water are P1 = 0.6117 kPa and T1 = 0.01ºC = 273.16 K (first line inTable A-4). Also, the enthalpy of sublimation of water at -30ºC is determined from Table A-8 to be 2838.4kJ/kg. Substituting,

kPa0.03799CCsub,

CCsub,

211

CCsub,

K15.273301

K16.2731

kJ/kg.K0.4615kJ/kg4.2838

kPa6117.0ln

11ln

P

P

TTRh

PP ig

The sublimation pressure of water at -30ºC is given in Table A-8 to be 0.03802 kPa. Then, the error involved in using Clapeyron-Clasius equation becomes

0.08%10003802.0

03799.003802.0orPercentErr

12-17

General Relations for du, dh, ds, cv, and cp

12-32C Yes, through the relation

PT

p

TT

Pc

2

2v

12-33 It is to be shown that the enthalpy of an ideal gas is a function of temperature only and that for anincompressible substance it also depends on pressure. Analysis The change in enthalpy is expressed as

dPT

TdTcdhP

Pv

v

For an ideal gas v = RT/P. Then,

Thus,dTcdh

PRT

TT

p

P0vvv

vv

To complete the proof we need to show that cp is not a function of P either. This is done with the help of the relation

PT

p

TT

Pc

2

2v

For an ideal gas,

Thus,

0

0)/(and2

2

T

P

PPP

Pc

TPR

TPR

Tvv

Therefore we conclude that the enthalpy of an ideal gas is a function of temperature only.For an incompressible substance v = constant and thus v/ T = 0. Then,

dPdTcdh p v

Therefore we conclude that the enthalpy of an incompressible substance is a function of temperature andpressure.

12-18

12-34 General expressions for u, h, and s for a gas that obeys the van der Waals equation of state foran isothermal process are to be derived.Analysis (a) For an isothermal process dT = 0 and the general relation for u reduces to

2

1

2

1

2

112

v

v v

v

v vv vv dP

TPTdP

TPTdTcuuu

T

T

The van der Waals equation of state can be expressed as

Thus,

22

2

vvvv

vvv

v

v

aab

RTb

RTPTPT

bR

TPa

bRTP

Substituting,

212

112

1 vvv

v

v

vadau

(b) The enthalpy change h is related to u through the relation

1122 vv PPuh

where

vvv

va

bRTP

Thus,

211

1

2

21122

11vvv

v

v

vvv a

bbRTPP

Substituting,

bbRTah

1

1

2

2

21

112v

v

v

v

vv

(c) For an isothermal process dT = 0 and the general relation for s reduces to

2

1

2

1

2

112

v

v v

v

v v

v vv dTPd

TPdT

Tc

sssT

T

Substituting ( P/ T)v = R/(v - b),

bb

Rdb

Rs1

2ln2

1 v

vv

v

v

v

12-19

12-35 General expressions for u, h, and s for a gas whose equation of state is P(v-a) = RT for an isothermal process are to be derived.Analysis (a) A relation for u is obtained from the general relation

2

1

2

112

v

v vv vdP

TPTdTcuuu

T

T

The equation of state for the specified gas can be expressed as

Thus,

0PPPa

RTPTPT

aR

TP

aRTP

v

vv

v

v

Substituting,2

1

T

TdTcu v

(b) A relation for h is obtained from the general relation

2

1

2

112

P

P P

T

TP dP

TvTdTchhh v

The equation of state for the specified gas can be expressed as

Thus,

aaPRT

TvT

PR

Ta

PRT

P

P

)(vvvv

vv

Substituting,

122

1

2

1

2

1

PPadTcdPadTchT

Tp

P

P

T

Tp

(c) A relation for s is obtained from the general relation

2

1

2

112

P

P P

T

T

p dPT

dTTc

sss v

Substituting ( v/ T)P = R/T,

1

2ln2

1

2

1

2

1 PP

RdTTc

dPPRdT

Tc

sT

T

pP

P P

T

T

p

For an isothermal process dT = 0 and these relations reduce to

1

212 lnand,,0

PP

RsPPahu

12-20

12-36 General expressions for ( u/ P)T and ( h/ v)T in terms of P, v, and T only are to be derived.Analysis The general relation for du is

vv

v dPTPTdTcdu

Differentiating each term in this equation with respect to P at T = constant yields

TTTT PP

PTPT

PP

TPT

Pu vvv

vv

0

Using the properties P, T, v, the cyclic relation can be expressed as

PTTP TPTP

PT

TP vvv

v vv

1

Substituting, we get

TPT PP

TT

Pu vv

The general relation for dh is

dPT

TdTcdhP

pv

v

Differentiating each term in this equation with respect to v at T = constant yields

TPTTPT

PT

TPPT

Thv

vv

vv

vv

v0

Using the properties v, T, P, the cyclic relation can be expressed as

vv vv

vv

PTP

TP

PT

T TPTP1

Substituting, we get

vvv

v PTTPh

TT

12-21

12-37 Expressions for the specific heat difference cp-cv for three substances are to be derived. Analysis The general relation for the specific heat difference cp - cv is

TPp

PT

Tccv

vv

2

(a) For an ideal gas Pv = RT. Then,

vvvv

vv

PRTPRTP

PR

TPRT

T

P

2

Substituting,

RRPTR

PRPTcc p vvv

2

(b) For a van der Waals gas RTbaP vv 2

. Then,

bT

Rba

aPR

baR

TbaPR

TP

vv

vv

vvv

vv

3

232

2

1211

Inverting,

bT

RbaT P

vv

vv

32

1

Also,322

2vvvvv

ab

RTPab

RTPT

Substituting,

32

2

3

22

1vv

vv

vva

bRT

bT

Rba

Tcc p

(c) For an incompressible substance v = constant and thus ( v / T)P = 0. Therefore, 0vcc p

12-22

12-38 The specific heat difference cp-cv for liquid water at 15 MPa and 80 C is to be estimated.Analysis The specific heat difference cp - cv is given as

TPp

PT

Tccv

vv

2

Approximating differentials by differences about the specified state,

Kkg/kJ0.3214K/kgmkPa3114.0

/kgm0.0010244)(0.0010199kPa10,000

K40/kgm0.0010105)(0.0010361

)K15.353(

MPa)10(20C)60100(

)K15.27380(

3

3

23

C80MPa10MPa20

2

MPa15

C60C100

C80

2

MPa15

TP

TPp

PT

Tcc

vv

vv

vv

v

12-39E The specific heat difference cp-cv for liquid water at 1000 psia and 150 F is to be estimated.Analysis The specific heat difference cp - cv is given as

TPp

PT

Tccv

vv

2


)ftpsia5.4039Btu(1R/lbmftpsia3081.0

/lbmft0.016317)(0.016267psia1000

R50/lbmft0.016177)(0.016427

R)609.67(

psia500)(1500F)125175(

)R67.459150(

33

3

23

F150psia500psia1500

2

psia1000

F125F175

F150

2

psia1000

RBtu/lbm0.0570

TP

TPp

PT

Tcc

vv

vv

vv

v

12-23

12-40 Relations for the volume expansivity and the isothermal compressibility for an ideal gas and for a gas whose equation of state is P(v-a) = RT are to be obtained. Analysis The volume expansivity and isothermal compressibility are expressed as

TP PTv

vv

v1and1

(a) For an ideal gas v = RT/P. Thus,

P11

T11

2 PPPRT

P

PR

PR

T

T

P

vv

vv

vv

(b) For a gas whose equation of state is v = RT/P + a,

vvv

vvv

vv

Pa

Pa

Pa

PRT

P

aPRTR

PR

PR

T

T

P

1

1

2

12-41 The volume expansivity and the isothermal compressibility of refrigerant-134a at 200 kPa and 30 C are to be estimated.Analysis The volume expansivity and isothermal compressibility are expressed as

TP PTv

vv

v1and1


1K003810.K20

/kgm0.11418)(0.12322/kgm0.11874

1

C)2040(11

3

3

kPa200

C20C40

kPa200 PPTvv

vv

v

and

1kPa004820.kPa60

/kgm0.13248)(0.09812/kgm0.11874

1

kPa180)(24011

3

3

C30

kPa180kPa240

C30 TTPvv

vv

v

12-24

The Joule-Thomson Coefficient

12-42C It represents the variation of temperature with pressure during a throttling process.

12-43C The line that passes through the peak points of the constant enthalpy lines on a T-P diagram is called the inversion line. The maximum inversion temperature is the highest temperature a fluid can be cooled by throttling.

12-44C No. The temperature may even increase as a result of throttling.

12-45C Yes.

12-46C No. Helium is an ideal gas and h = h(T) for ideal gases. Therefore, the temperature of an ideal gas remains constant during a throttling (h = constant) process.

12-47 The equation of state of a gas is given to be P(v-a) = RT. It is to be determined if it is possible to cool this gas by throttling. Analysis The equation of state of this gas can be expressed as

PR

Ta

PRT

P

vv

Substituting into the Joule-Thomson coefficient relation,

0111

pppPp caa

cPRT

cTT

cvvv

vv

Therefore, this gas cannot be cooled by throttling since is always a negative quantity.

12-48 Relations for the Joule-Thompson coefficient and the inversion temperature for a gas whose equation of state is (P+a/v2) v = RT are to be obtained. Analysis The equation of state of this gas can be expressed as

222322212

v

vvv

v

vv

vvv

v

RaRTT

RaaP

Ra

RTaP

RT

P

Substituting into the Joule-Thomson coefficient relation,

)2(2

211 2

vv

vv

vv

vRTac

aaRT

RTcT

Tc ppPp

The temperature at = 0 is the inversion temperature,

00)2(

2v

vvRTac

a

p

Thus the line of v = 0 is the inversion line. Since it is not physically possible to have v = 0, this gas does not have an inversion line.

12-25

12-49 The Joule-Thompson coefficient of steam at two states is to be estimated.Analysis (a) The enthalpy of steam at 3 MPa and 300 C is h = 2994.3 kJ/kg. Approximating differentials by differences about the specified state, the Joule-Thomson coefficient is expressed as

kJ/kg3.2994hh PT

PT

Considering a throttling process from 3.5 MPa to 2.5 MPa at h = 2994.3 kJ/kg, the Joule-Thomsoncoefficient is determined to be

C/MPa12.3MPa)5.2(3.5

C)2943.306(MPa)5.2(3.5

kJ/kg3.2994

MPa2.5MPa3.5

h

TT

(b) The enthalpy of steam at 6 MPa and 500 C is h = 3423.1 kJ/kg. Approximating differentials bydifferences about the specified state, the Joule-Thomson coefficient is expressed as

kJ/kg1.3423hh PT

PT


C/MPa4.9MPa5.0)(7.0

C)1.4958.504(MPa5.0)(7.0

kJ/kg1.3423

MPa5.0MPa7.0

h

TT

12-50E [Also solved by EES on enclosed CD] The Joule-Thompson coefficient of nitrogen at two states is to be estimated.Analysis (a) The enthalpy of nitrogen at 200 psia and 500 R is, from EES, h = -10.564 Btu/lbm. Note that in EES, by default, the reference state for specific enthalpy and entropy is 0 at 25ºC (77ºF) and 1 atm.Approximating differentials by differences about the specified state, the Joule-Thomson coefficient is expressed as

Btu/lbm564.10hh PT

PT

Considering a throttling process from 210 psia to 190 psia at h = -10.564 Btu/lbm, the Joule-Thomsoncoefficient is determined to be

R/psia0.0297psia)210(190

R500.296)(499.703psia210)(190

Btu/lbm564.10

psia210psia190

h

TT

(b) The enthalpy of nitrogen at 2000 psia and 400 R is, from EES, h = -55.321 Btu/lbm. Approximatingdifferentials by differences about the specified state, the Joule-Thomson coefficient is expressed as

Btu/lbm321.55hh PT

PT

Considering a throttling process from 2010 psia to 1990 psia at h = -55.321 Btu/lbm, the Joule-Thomsoncoefficient is determined to be

R/psia0.0213psia2010)(1990

R400.213)-(399.786psia2010)(1990

Btu/lbm321.55

psia2001psia1999

h

TT

12-26

12-51E EES Problem 12-50E is reconsidered. The Joule-Thompson coefficient for nitrogen over the pressure range 100 to 1500 psia at the enthalpy values 100, 175, and 225 Btu/lbm is to be plotted. Analysis The problem is solved using EES, and the results are tabulated and plotted below.

Gas$ = 'Nitrogen' {P_ref=200 [psia] T_ref=500 [R] P= P_ref}h=50 [Btu/lbm] {h=enthalpy(Gas$, T=T_ref, P=P_ref)}dP = 10 [psia] T = temperature(Gas$, P=P, h=h) P[1] = P + dP P[2] = P - dP T[1] = temperature(Gas$, P=P[1], h=h) T[2] = temperature(Gas$, P=P[2], h=h) Mu = DELTAT/DELTAP "Approximate the differential by differences about the state at h=const."DELTAT=T[2]-T[1]DELTAP=P[2]-P[1]

h = 100 Btu/lbmP [psia] μ [R/psia]

100 0.003675275 0.003277450 0.002899625 0.00254800 0.002198975 0.001871

1150 0.0015581325 0.0012581500 0.0009699

0 200 400 600 800 1000 1200 1400 1600-0.006

-0.005

-0.004

-0.003

-0.002

-0.001

0

0.001

0.002

0.003

0.004

P [psia]

[R/psia]

h = 100 Btu/lbm

h = 175 Btu/lbm

h = 225 Btu/lbm

12-27

12-52 The Joule-Thompson coefficient of refrigerant-134a at a specified state is to be estimated.Analysis The enthalpy of refrigerant-134a at 0.7 MPa and T = 50 C is h = 288.53 kJ/kg. Approximatingdifferentials by differences about the specified state, the Joule-Thomson coefficient is expressed as

kJ/kg53.288hh PT

PT


C/MPa18.1MPa0.6)(0.8

C.19)48(51.81MPa0.6)(0.8

kJ/kg53.288

MPa0.6MPa0.8

h

TT

12-53 Steam is throttled slightly from 1 MPa and 300 C. It is to be determined if the temperature of the steam will increase, decrease, or remain the same during this process. Analysis The enthalpy of steam at 1 MPa and T = 300 C is h = 3051.6 kJ/kg. Now consider a throttling process from this state to 0.8 MPa, which is the next lowest pressure listed in the tables. The temperatureof the steam at the end of this throttling process will be

C52.297kJ/kg6.3051

MPa8.02T

hP

Therefore, the temperature will decrease.

12-28

The h, u, and s of Real Gases

12-54C It is the variation of enthalpy with pressure at a fixed temperature.

12-55C As PR approaches zero, the gas approaches ideal gas behavior. As a result, the deviation from ideal gas behavior diminishes.

12-56C So that a single chart can be used for all gases instead of a single particular gas.

12-57 The enthalpy of nitrogen at 175 K and 8 MPa is to be determined using data from the ideal-gas nitrogen table and the generalized enthalpy departure chart. Analysis (a) From the ideal gas table of nitrogen (Table A-18) we read

kg/kmol)013.28(kJ/kmol8.50832NMh kJ/kg181.48

N2175 K 8 MPa

at the specified temperature. This value involves 44.4% error. (b) The enthalpy departure of nitrogen at the specified state is determined from the generalized chart to be

and 6.1)(

360.239.38

387.12.126

175

cr

,ideal

cr

cr

TRhh

Z

PPP

TTT

u

PTh

R

R

Thus,

or,

)error%1.3(kg/kmol28.013kJ/kmol3405.0

kJ/kmol0.34052.126314.86.18.5083crideal

kJ/kg121.6Mhh

TRZhh uh

12-29

12-58E The enthalpy of nitrogen at 400 R and 2000 psia is to be determined using data from the ideal-gasnitrogen table and the generalized enthalpy departure chart.Analysis (a) From the ideal gas table of nitrogen (Table A-18E) we read

)lbm/lbmol28(Btu/lbmol0.27772NMh Btu/lbm99.18

N2400 R

2000 psia

at the specified temperature. This value involves 44.2% error. (b) The enthalpy departure of nitrogen at the specified state is determined from the generalized chart to be (Fig. A-29)

and 18.1)(

065.4492

2000

761.11.227

400,ideal

cr

cr

cru

PTh

R

R

TRhh

Z

PPP

TTT

Thus,

or,

error)(54.9%lbm/lbmol28

Btu/lbmol2244.8

Btu/lbmol8.22441.227986.118.10.2777crideal

Btu/lbm80.17Mhh

TRZhh uh

12-59 The errors involved in the enthalpy and internal energy of CO2 at 350 K and 10 MPa if it is assumedto be an ideal gas are to be determined.Analysis (a) The enthalpy departure of CO2 at the specified state is determined from the generalized chart to be (Fig. A-29)

and 5.1)(

353.139.7

10

151.12.304

350

cr

,ideal

cr

cr

TRhh

Z

PPP

TTT

u

PTh

R

R

CO2350 K

10 MPaThus,

and,

%.557,7

557,7351,11)(Error

kJ/kmol557,72.304314.85.1351,11

,ideal

crideal

250h

hh

TRZhh

PT

uh

(b) At the calculated TR and PR the compressibility factor is determined from the compressibility chart tobe Z = 0.65. Then using the definition of enthalpy, the internal energy is determined to be

and,

%9.48666,5

666,5439,8Error

kJ/kmol666,5350314.865.07557

ideal

uuu

TZRhPhu uv

12-30

12-60 The enthalpy and entropy changes of nitrogen during a process are to be determined assuming idealgas behavior and using generalized charts.Analysis (a) Using data from the ideal gas property table of nitrogen (Table A-18),

and

K/.6

12ln314.8289.183562.193ln)(

/537,69306)(

1

212ideal12

ideal1,ideal2,ideal12

kmolkJ5104

kmolkJ2769

PP

Rssss

hhhh

u

(b) The enthalpy and entropy departures of nitrogen at the specified states are determined from the generalized charts to be (Figs. A-29, A-30)

25.0and6.0770.1

39.36

783.12.126

225

11

cr

11

cr

11

sh

R

R

ZZ

PP

P

TT

T

and

15.0and4.0540.2

39.312

536.22.126

320

22

cr

22

cr

22

sh

R

R

ZZ

PP

P

TT

T

Substituting,

Kkmol/kJ341.5

kmol/kJ2979

510.415.025.0314.8)()(

27694.06.02.126314.8)()(

ideal122112

ideal122112

ssZZRss

hhZZTRhh

ssu

hhcru

12-31

12-61 The enthalpy and entropy changes of CO2 during a process are to be determined assuming ideal gas behavior and using generalized charts.

280 K 12 MPa

CO2250 K 7 MPa

Analysis (a) Using data from the ideal gas propertytable of CO2 (Table A-20),

Thus,

KkJ/kg0.0100

kJ/kg24.32

kg/kmol44kJ/kmol0.442)(

)(

kg/kmol44kJ/kmol1,070)(

)(

Kkmol/kJ442.07

12ln314.8337.207376.211ln)(

kmol/kJ070,1627,7697,8)(

ideal12ideal12

ideal12ideal12

1

212ideal12


Mss

ss

Mhh

hh

PP

Rssss

hhhh

u

(b) The enthalpy and entropy departures of CO2 at the specified states are determined from the generalizedcharts to be (Figs. A-29, A-30)

and

2.4and0.5624.1

39.712

920.02.304

280

3.5and5.5947.0

39.77

822.02.304

250

22

cr

22

cr

22

11

cr

11

cr

11

sh

R

R

sh

R

R

ZZ

PP

P

TT

T

ZZ

PP

P

TT

T

Thus,

KkJ/kg0.198

kg/kJ05.53

010.02.43.51889.0)()(

32.240.55.52.3041889.0)()(

ideal122112

ideal1221cr12

ssZZRss

hhZZRThh

ss

hh

12-32

12-62 Methane is compressed adiabatically by a steady-flow compressor. The required power input to thecompressor is to be determined using the generalized charts.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis The steady-flow energy balance equation for this compressor can be expressed as

12inC,

21inC,

outin

(steady)0systemoutin 0

hhmW

hmhmW

EE

EEE

10 MPa 110 C

CH4 = 0.55 kg/s m·The enthalpy departures of CH4 at the specified states are determined

from the generalized charts to be (Fig. A-29)

and

50.0155.2

64.410

00.21.191

383

21.0431.0

64.42

376.11.191

263

2

cr

22

cr

22

1

cr

11

cr

11

h

R

R

h

R

R

Z

PP

P

TT

T

Z

PP

P

TT

T2 MPa -10 C

Thus,

kg/kJ7.241101102537.250.021.01.1915182.0)()( ideal1221cr12 hhZZRThh hh

Substituting,

kW133kJ/kg241.7kg/s0.55inC,W

12-33

12-63 [Also solved by EES on enclosed CD] Propane is compressed isothermally by a piston-cylinderdevice. The work done and the heat transfer are to be determined using the generalized charts.Assumptions 1 The compression process is quasi-equilibrium. 2 Kinetic and potential energy changes are negligible.Analysis (a) The enthalpy departure and the compressibility factors of propane at the initial and the finalstates are determined from the generalized charts to be (Figs. A-29, A-15)

and

50.0and8.1939.0

26.44

008.1370373

92.0and28.0235.0

26.41

008.1370373

22

cr

22

cr

22

11

cr

11

cr

11

ZZ

PP

P

TT

T

ZZ

PP

P

TT

T

h

R

R

h

R

R

Q

Propane1 MPa 100 C

Treating propane as a real gas with Zavg = (Z1+Z2)/2 = (0.92 + 0.50)/2 = 0.71, constantavg CRTZZRTPv

Then the boundary work becomes

kJ/kg99.640.9210.50lnK373KkJ/kg0.188571.0

ln//

lnln21

12ave

11

22avg

1

22

1

2

1in, PZ

PZRTZ

PRTZPRTZ

RTZCdCdPwb v

vv

vv

Also,

kg/kJ5.7637392.03735.01885.0106)()(

kg/kJ10608.128.03701885.0)()(

11221212

ideal122112

TZTZRhhuu

hhZZRThh hhcr

Then the heat transfer for this process is determined from the closed system energy balance to be

kJ/kg176.1kg/kJ1.1766.995.76 outin,12in

12in,in

systemoutin

qwuuquuuwq

EEE

b

b

12-34

12-64 EES Problem 12-63 is reconsidered. This problem is to be extended to compare the solutions basedon the ideal gas assumption, generalized chart data and real fluid (EES) data. Also, the solution is to beextended to carbon dioxide, nitrogen and methane.Analysis The problem is solved using EES, and the solution is given below.

Procedure INFO(Name$, T[1] : Fluid$, T_critical, p_critical)If Name$='Propane' then

T_critical=370 ; p_critical=4620 ; Fluid$='C3H8'; goto 10 endifIf Name$='Methane' then

T_critical=191.1 ; p_critical=4640 ; Fluid$='CH4'; goto 10 endifIf Name$='Nitrogen' then

T_critical=126.2 ; p_critical=3390 ; Fluid$='N2'; goto 10 endifIf Name$='Oxygen' then

T_critical=154.8 ; p_critical=5080 ; Fluid$='O2'; goto 10 endifIf Name$='CarbonDioxide' then

T_critical=304.2 ; p_critical=7390 ; Fluid$='CO2' ; goto 10 endifIf Name$='n-Butane' then

T_critical=425.2 ; p_critical=3800 ; Fluid$='C4H10' ; goto 10 endif

10: If T[1]<=T_critical thenCALL ERROR('The supplied temperature must be greater than the critical temperature for the fluid. A value of XXXF1 K was supplied',T[1])endif

end

{"Data from the Diagram Window"T[1]=100+273.15p[1]=1000p[2]=4000Name$='Propane'Fluid$='C3H8' }

Call INFO(Name$, T[1] : Fluid$, T_critical, p_critical)R_u=8.314M=molarmass(Fluid$)R=R_u/M

"****** IDEAL GAS SOLUTION ******""State 1"h_ideal[1]=enthalpy(Fluid$, T=T[1]) "Enthalpy of ideal gas"s_ideal[1]=entropy(Fluid$, T=T[1], p=p[1]) "Entropy of ideal gas"u_ideal[1]=h_ideal[1]-R*T[1] "Internal energy of ideal gas""State 2"h_ideal[2]=enthalpy(Fluid$, T=T[2]) "Enthalpy of ideal gas"s_ideal[2]=entropy(Fluid$, T=T[2], p=p[2]) "Entropy of ideal gas"u_ideal[2]=h_ideal[2]-R*T[2] "Internal energy of ideal gas"

12-35

"Work is the integral of p dv, which can be done analytically."w_ideal=R*T[1]*Ln(p[1]/p[2])

"First Law - note that u_ideal[2] is equal to u_ideal[1]"q_ideal-w_ideal=u_ideal[2]-u_ideal[1]

"Entropy change"DELTAs_ideal=s_ideal[2]-s_ideal[1]

"***** COMPRESSABILITY CHART SOLUTION ******""State 1"Tr[1]=T[1]/T_criticalpr[1]=p[1]/p_criticalZ[1]=COMPRESS(Tr[1], Pr[1]) DELTAh[1]=ENTHDEP(Tr[1], Pr[1])*R*T_critical "Enthalpy departure"h[1]=h_ideal[1]-DELTAh[1] "Enthalpy of real gas using charts"u[1]=h[1]-Z[1]*R*T[1]

"Internal energy of gas using charts"DELTAs[1]=ENTRDEP(Tr[1], Pr[1])*R "Entropy departure"s[1]=s_ideal[1]-DELTAs[1] "Entropy of real gas using charts""State 2"T[2]=T[1]Tr[2]=Tr[1]pr[2]=p[2]/p_criticalZ[2]=COMPRESS(Tr[2], Pr[2]) DELTAh[2]=ENTHDEP(Tr[2], Pr[2])*R*T_critical "Enthalpy departure"DELTAs[2]=ENTRDEP(Tr[2], Pr[2])*R "Entropy departure"h[2]=h_ideal[2]-DELTAh[2] "Enthalpy of real gas using charts"s[2]=s_ideal[2]-DELTAs[2] "Entropy of real gas using charts"u[2]=h[2]-Z[2]*R*T[2] "Internal energy of gas using charts"

"Work using charts - note use of EES integral function to evaluate the integral of p dv."w_chart=Integral(p,v,v[1],v[2])"We need an equation to relate p and v in the above INTEGRAL function. "p*v=COMPRESS(Tr[2],p/p_critical)*R*T[1] "To specify relationship between p and v""Find the limits of integration"p[1]*v[1]=Z[1]*R*T[1] "to get v[1], the lower bound"p[2]*v[2]=Z[2]*R*T[2] "to get v[2], the upper bound"

"First Law - note that u[2] is not equal to u[1]"q_chart-w_chart=u[2]-u[1]

"Entropy Change"DELTAs_chart=s[2]-s[1]

"***** SOLUTION USING EES BUILT-IN PROPERTY DATA *****""At state 1"u_ees[1]=intEnergy(Name$,T=T[1],p=p[1])s_ees[1]=entropy(Name$,T=T[1],p=p[1])"At state 2"u_ees[2]=IntEnergy(Name$,T=T[2],p=p[2])s_ees[2]=entropy(Name$,T=T[2],p=p[2])

"Work using EES built-in properties- note use of EES Integral funcion to evaluate the integral of pdv."w_ees=integral(p_ees, v_ees, v_ees[1],v_ees[2]) "The following equation relates p and v in the above INTEGRAL"

12-36

p_ees=pressure(Name$,T=T[1], v=v_ees) "To specify relationship between p and v"

"Find the limits of integration"v_ees[1]=volume(Name$, T=T[1],p=p[1]) "to get lower bound"v_ees[2]=volume(Name$, T=T[2],p=p[2]) "to get upper bound"

"First law - note that u_ees[2] is not equal to u_ees[1]"q_ees-w_ees=u_ees[2]-u_ees[1]

"Entropy change"DELTAs_ees=s_ees[2]-s_ees[1]

"Note: In all three solutions to this problem we could have calculated the heat transfer by q/T=DELTA_s since T is constant. Then the first law could have been used to find the work.The use of integral of p dv to find the work is a more fundamental approach and can be used if T is not constant."

SOLUTION

DELTAh[1]=16.48 [kJ/kg] DELTAh[2]=91.96 [kJ/kg] DELTAs[1]=0.03029 [kJ/kg-K] DELTAs[2]=0.1851 [kJ/kg-K] DELTAs_chart=-0.4162 [kJ/kg-K] DELTAs_ees=-0.4711 [kJ/kg-K] DELTAs_ideal=-0.2614 [kJ/kg-K] Fluid$='C3H8'h[1]=-2232 [kJ/kg] h[2]=-2308 [kJ/kg] h_ideal[1]=-2216 [kJ/kg] h_ideal[2]=-2216 [kJ/kg] M=44.1Name$='Propane'p=4000p[1]=1000 [kPa] p[2]=4000 [kPa] pr[1]=0.2165pr[2]=0.8658p_critical=4620 [kPa] p_ees=4000q_chart=-155.3 [kJ/kg] q_ees=-175.8 [kJ/kg] q_ideal=-97.54 [kJ/kg] R=0.1885 [kJ/kg-K] R_u=8.314 [kJ/mole-K]s[1]=6.073 [kJ/kg-K]

s[2]=5.657 [kJ/kg-K] s_ees[1]=2.797 [kJ/kg-K] s_ees[2]=2.326 [kJ/kg-K] s_ideal[1]=6.103 [kJ/kg-K] s_ideal[2]=5.842 [kJ/kg-K] T[1]=373.2 [K] T[2]=373.2 [K] Tr[1]=1.009Tr[2]=1.009T_critical=370 [K] u[1]=-2298 [kJ/kg] u[2]=-2351 [kJ/kg] u_ees[1]=688.4 [kJ/kg] u_ees[2]=617.1 [kJ/kg] u_ideal[1]=-2286 [kJ/kg] u_ideal[2]=-2286 [kJ/kg] v=0.01074v[1]=0.06506 [m^3/kg] v[2]=0.01074 [m^3/kg] v_ees=0.009426v_ees[1]=0.0646 [m^3/kg] v_ees[2]=0.009426 [m^3/kg] w_chart=-101.9 [kJ/kg] w_ees=-104.5 [kJ/kg] w_ideal=-97.54 [kJ/kg] Z[1]=0.9246Z[2]=0.6104

12-37

12-65E Propane is compressed isothermally by a piston-cylinder device. The work done and the heattransfer are to be determined using the generalized charts.Assumptions 1The compression process is quasi-equilibrium. 2 Kinetic and potential energy changes are negligible. 3 The device is well-insulated and thus heat transfer is negligibleAnalysis (a) The enthalpy departure and the compressibility factors of propane at the initial and the finalstates are determined from the generalized charts to be (Figs. A-29, A-15)

and

22.0and2.4297.1

617800

991.09.665

660

88.0and37.0324.0

617200

991.09.665

660

22

cr

22

cr

22

11

cr

11

cr

11

ZZ

PP

P

TT

T

ZZ

PP

P

TT

T

h

R

R

h

R

R

Q

Propane200 psia 200 F

Treating propane as a real gas with Zavg = (Z1+Z2)/2 = (0.88 + 0.22)/2 = 0.55, constantavg CRTZZRTPv

Then the boundary work becomes

Btu/lbm45.380088.020022.0lnR660RBtu/lbm0.0450455.0

ln//

lnln21

12avg

11

22avg

1

22

1

2

1in, PZ

PZRTZ

PRTZPRTZ

RTZCdCdPwb v

vv

vv

Also,

lbm/Btu3.9566088.066022.004504.09.114)()(

lbm/Btu9.11402.437.09.66504504.0)()(

11221212

0ideal122112

TZTZRhhuu

hhZZRThh hhcr

Then the heat transfer for this process is determined from the closed system energy balance equation to be

lbmBtu6140 /.Btu/lbm6.1403.453.95 outin,12in

12in,in

systemoutin

qwuuquuuwq

EEE

b

b

12-38

12-66 Propane is compressed isothermally by a piston-cylinder device. The exergy destruction associated with this process is to be determined.Assumptions 1The compression process is quasi-equilibrium. 2 Kinetic and potential energy changes are negligible.Properties The gas constant of propane is R = 0.1885 kJ/kg.K (Table A-1). Analysis The exergy destruction is determined from its definition where the entropy generation is determined from an entropy balance on the contents of the cylinder. It gives

x Tdestroyed gen0s

)(=)(surr

out12gen12gen

surr,

out

systemgenoutin

Tq

sssssmSTQ

SSSS

b

where

KkJ/kg261.014ln1885.00lnln)(

)()(

1

20

1

2ideal12

ideal122112sys

PP

RTT

css

ssZZRsss

p

ss

and

5.1939.0

26.44

008.1370373

21.0235.0

26.41

008.1370373

2

cr

22

cr

22

1

cr

11

cr

11

s

R

R

s

R

R

Z

PP

P

TT

T

Z

PP

P

TT

T

Thus,KkJ/kg.5040261.05.121.01885.0)()( ideal122112sys ssZZRsss ss

and

kgkJ423 /.KkJ/kgK303

kJ/kg176.10.504K303)(

surr

out120gen0destroyed T

qssTsTx

12-39

12-67 Carbon dioxide passes through an adiabatic nozzle. The exit velocity is to be determined using thegeneralized enthalpy departure chart. Assumptions 1Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3The nozzle is adiabatic and thus heat transfer is negligibleProperties The gas constant of CO2 is 0.1889 kJ/kg.K (Table A-1). Analysis The steady-flow energy balanceequation for this nozzle can be expressed as

P2 = 2 MPaT2 = 350 K

CO2

)(2

)2/()2/(

0

212

222

0211

outin

(steady)0systemoutin

hhV

VhVh

EE

EEE P1 = 8 MPaT1 = 450 K

The enthalpy departures of CO2 at the specified states are determined from the generalized enthalpydeparture chart to be

and

20.027.0

39.72

15.12.304

350

55.008.1

39.78

48.12.304

450

2

cr

22

cr

22

1

cr

11

cr

11

h

R

R

h

R

R

Z

PP

P

TT

T

Z

PP

P

TT

T

Thus,

kJ/kg8.7344/483,15351,112.055.02.3041889.0)()( ideal1221cr12 hhZZRThh hh

Substituting,

m/s384kJ/kg1

/sm1000kJ/kg73.82

22

2V

12-40

12-68 EES Problem 12-67 is reconsidered. the exit velocity to the nozzle assuming ideal gas behavior, thegeneralized chart data, and EES data for carbon dioxide are to be compared.Analysis The problem is solved using EES, and the results are given below.

Procedure INFO(Name$: Fluid$, T_critical, p_critical) If Name$='CarbonDioxide' then

T_critical=304.2 ; p_critical=7390 ; Fluid$='CO2'endif

END

T[1]=450 [K] P[1]=8000 [kPa] P[2]=2000 [kPa] T[2]=350 [K] Name$='CarbonDioxide'

Call INFO(Name$: Fluid$, T_critical, P_critical) R_u=8.314M=molarmass(Fluid$)R=R_u/M

"****** IDEAL GAS SOLUTION ******""State 1 nd 2"h_ideal[1]=enthalpy(Fluid$, T=T[1]) "Enthalpy of ideal gas"h_ideal[2]=enthalpy(Fluid$, T=T[2]) "Enthalpy of ideal gas""Exit velocity:"V_2_ideal=SQRT(2*(h_ideal[1]-h_ideal[2])*convert(kJ/kg,m^2/s^2))"[m/s]"

"***** COMPRESSABILITY CHART SOLUTION ******""State 1"Tr[1]=T[1]/T_criticalPr[1]=P[1]/P_criticalDELTAh[1]=ENTHDEP(Tr[1], Pr[1])*R*T_critical "Enthalpy departure"h[1]=h_ideal[1]-DELTAh[1] "Enthalpy of real gas using charts"

"State 2"Tr[2]=T[2]/T_criticalPr[2]=P[2]/P_criticalDELTAh[2]=ENTHDEP(Tr[2], Pr[2])*R*T_critical "Enthalpy departure"h[2]=h_ideal[2]-DELTAh[2] "Enthalpy of real gas using charts"

"Exit velocity:"V_2_EnthDep=SQRT(2*(h[1]-h[2])*convert(kJ/kg,m^2/s^2)) "[m/s]"

"***** SOLUTION USING EES BUILT-IN PROPERTY DATA *****""At state 1 and 2"h_ees[1]=enthalpy(Name$,T=T[1],P=P[1])h_ees[2]=enthalpy(Name$,T=T[2],P=P[2])

"Exit velocity:"V_2_ees=SQRT(2*(h_ees[1]-h_ees[2])*convert(kJ/kg,m^2/s^2))"[m/s]"

12-41

SOLUTIONDELTAh[1]=34.19 [kJ/kg] DELTAh[2]=13.51 [kJ/kg] Fluid$='CO2'h[1]=-8837 [kJ/kg] h[2]=-8910 [kJ/kg] h_ees[1]=106.4 [kJ/kg] h_ees[2]=31.38 [kJ/kg] h_ideal[1]=-8803 [kJ/kg] h_ideal[2]=-8897 [kJ/kg] M=44.01 [kg/kmol]Name$='CarbonDioxide'P[1]=8000 [kPa] P[2]=2000 [kPa]

Pr[1]=1.083Pr[2]=0.2706P_critical=7390 [kPa] R=0.1889 [kJ/kg-K] R_u=8.314 [kJ/kmol-K]T[1]=450 [K] T[2]=350 [K] Tr[1]=1.479Tr[2]=1.151T_critical=304.2 [K] V_2_ees=387.4 [m/s] V_2_EnthDep=382.3 [m/s] V_2_ideal=433 [m/s]

12-69 A paddle-wheel placed in a well-insulated rigid tank containing oxygen is turned on. The final pressure in the tank and the paddle-wheel work done during this process are to be determined.Assumptions 1The tank is well-insulated and thus heat transfer is negligible. 2 Kinetic and potential energychanges are negligible.Properties The gas constant of O2 is R = 0.2598 kJ/kg.K (Table A-1). Analysis (a) The compressibility factor of oxygen at theinitial state is determined from the generalized chart to be O2

220 K 10 MPa 15.1and80.0

97.108.5

10

42.18.154

220

11

cr

11

cr

11

h

R

R

ZZ

PP

P

TT

T

Then,

kg5.17/kgm0.00457

m0.08

/kgm0.00457kPa10,000

K)K)(220/kgmkPa98(0.8)(0.25

3

3

1

33

1

vV

vv

m

ZRTP

The specific volume of oxygen remains constant during this process, v2 = v1. Thus,

kPa12,19050804.2

4.20.187.0

577.0kPa)(5080/K)K)(154.8/kgmkPa(0.2598

/kgm0.00457/

615.18.154

250

cr22

2

2

2

3

3

crcr

22

cr

22

PPP

PZZ

PRT

TT

T

R

R

hR

R

vv

(b) The energy balance relation for this closed system can be expressed as

)()()(

112212112212in

12in

systemoutin

TZTZRhhmPPhhmWuumUW

EEE

vv

wherekJ/kg25.3332/)64047275()115.1)(8.154)(2598.0(

)()( ideal1221cr12 hhZZRThh hh

Substituting,kJ393K2200.802500.87K)kJ/kg(0.259833.25kg)5.17(inW

12-42

12-70 The heat transfer and entropy changes of CO2 during a process are to be determined assuming idealgas behavior, using generalized charts, and real fluid (EES) data.Analysis The temperature at the final state is

K2984MPa1MPa8K)273100(

1

212 P

PTT

Using data from the ideal gas property table of CO2 (Table A-20),

kJ/kg9.3386kg/kmol44

kJ/kmol149,024)()(

Kkmol/kJ115.9418ln314.8367.222770.333ln)(

kmol/kJ024,149269,,12293,161)(

ideal12ideal12

1

212ideal12


Mhh

hh

PP

Rssss

hhhh

u

The heat transfer is determined from an energy balance noting that there is no work interaction

kJ/kg2893.7373)-984kJ/kg.K)(2(0.1889-kJ/kg9.3386)()()( 12ideal12ideal12ideal TTRhhuuq

”

The entropy change is

kJ/kg.K2.1390kg/kmol44

kJ/kmol115.94)()( ideal12

ideal12ideal Mss

sss

The compressibility factor and the enthalpy and entropy departures of CO2 at the specified states are determined from the generalized charts to be (we used EES)

and

002685.0and1144.0,009.1083.1

39.78

813.92.304

2985

05987.0and1028.0,976.0135.0

39.71

226.12.304

373

222

cr

22

cr

22

111

cr

11

cr

11

sh

R

R

sh

R

R

ZZZ

PP

P

TT

T

ZZZ

PP

P

TT

T

Thus,

kJ/kg.K2.151

kJ/kg2935.9

1390.2)002685.0(05987.01889.0

)()()(

)3732887)(1889.0)(976.0()1028.01144.0)(2.304)(1889.0(9.3386)()()(

ideal1221chart12chart

12112crideal1212chart

ssZZRsss

TTRZZZRThhuuq

ss

hh

Note that the temperature at the final state in this case was determined from

K2888009.1976.0

MPa1MPa8K)273100(

2

1

1

212 Z

ZPP

TT

The solution using EES built-in property data is as follows:

kJ/kg.K2464.0

kJ/kg614.8

/kgm06885.0

MPa1K373

1

1

31

1

1

s

uPT

v

kJ/kg.K85.1

kJ/kg2754

K2879

/kgm06885.0

MPa8

2

2

2

312

2

s

u

TP

vv

Then

kJ/kg.K2.097

kJ/kg2763

)2464.0(85.1)(

)614.8(2754

12EES12EES

12EES

sssss

uuq

12-43

Review Problems

12-71 For 0, it is to be shown that at every point of a single-phase region of an h-s diagram, the slopeof a constant-pressure line is greater than the slope of a constant-temperature line, but less than the slope of a constant-volume line.Analysis It is given that > 0.

Using the Tds relation:dsdPT

dsdhdPdsTdh vv

(1) P = constant: Tsh

P

(2) T = constant:TT s

PTsh

v

But the 4th Maxwell relation:PT

TsP

v

Substituting: 1TTTsh

PT vv

Therefore, the slope of P = constant lines is greater than the slope of T = constant lines.

(3) v = constant: )(asPT

sh

vv

v

From the ds relation: vv

v dTPdT

Tc

ds

Divide by dP holding v constant: )(or bTP

cT

sP

PT

Tc

Ps

vvvv

v

v

Using the properties P, T, v, the cyclic relation can be expressed as

)(11 cPTT

PP

TTP

TPTP vv

vvv

v vv

where we used the definitions of and . Substituting (b) and (c) into (a),

TcTT

sPT

sh

vvv

vv

Here is positive for all phases of all substances. T is the absolute temperature that is also positive, so is cv. Therefore, the second term on the right is always a positive quantity since is given to be positive.Then we conclude that the slope of P = constant lines is less than the slope of v = constant lines.

12-44

12-72 Using the cyclic relation and the first Maxwell relation, the other three Maxwell relations are to beobtained.Analysis (1) Using the properties P, s, v, the cyclic relation can be expressed as

1sP P

ssP v

vv

Substituting the first Maxwell relation,vv s

PT

s,

PsPssPs sPTs

PT

PsT v

vv

vv11

(2) Using the properties T, v, s, the cyclic relation can be expressed as

1v

vv T

ss

T

Ts

Substituting the first Maxwell relation,vv s

PT

s,

vvvv vvv

TPs

sTP

Ts

ssP

TTT11

(3) Using the properties P, T, v, the cyclic relation can be expressed as

1TP P

TTP v

vv

Substituting the third Maxwell relation,vv T

Ps

T,

PTPTTPT TPsT

Ps

PTs v

vv

vv11

12-73 It is to be shown that the slope of a constant-pressure line on an h-s diagram is constant in thesaturation region and increases with temperature in the superheated region. Analysis For P = constant, dP = 0 and the given relation reduces to dh = Tds, which can also be expressed as

Tsh

Ph

P = const.

s

Thus the slope of the P = constant lines on an h-s diagram is equal tothe temperature.(a) In the saturation region, T = constant for P = constant lines, and theslope remains constant.(b) In the superheat region, the slope increases with increasing temperature since the slope is equal temperature.

12-45

12-74 The relations for u, h, and s of a gas that obeys the equation of state (P+a/v2)v = RT for an isothermal process are to be derived.Analysis (a) For an isothermal process dT = 0 and the general relation for u reduces to

2

1

2

1

2

112

v

v v

v

v vv vv dP

TPTdP

TPTdTcuuu

T

T

For this gas the equation of state can be expressed as

vvv v

RTPaRTP

2

Thus,

Substituting,

212

22

112

1 vvv

v

vvvv

v

v

v

adau

aaRTRTPTPT

(b) The enthalpy change h is related to u through the relation

where

vv

vv

aRTP

PPuh 1122

Thus,

Substituting,

21

21121122

112

11

vv

vvvvvv

ah

aaRTaRTPP

(c) For an isothermal process dT = 0 and the general relation for s reduces to

2

1

2

1

2

112

v

v v

v

v v

v vv dTPd

TPdT

Tc

sssT

T

Substituting ( P/ T)v = R/v,2

1 1

2lnv

v v

vv

vRdRs

12-46

12-75 It is to be shown that

vv

vTP

TTc

s and

Psp TT

PTc v

Analysis Using the definition of cv ,

vvvv T

PPsT

TsTc

Substituting the first Maxwell relationsTP

s v

v

,

vv

vTP

TTc

s

Using the definition of cp,

PPPp T

sTTsTc v

v

Substituting the second Maxwell relation sP T

Psv

,

Psp TT

PTc v

12-76 The Cp of nitrogen at 300 kPa and 400 K is to be estimated using the relation given and itsdefinition, and the results are to be compared to the value listed in Table A-2b.Analysis (a) We treat nitrogen as an ideal gas with R = 0.297 kJ/kg·K and k = 1.397. Note that PT-k/(k-1) = C = constant for the isentropic processes of ideal gases. The cp relation is given as

)1(111)1/()1/(1)1/()1/(

kTkPTPT

kkCT

kk

TPCTP

PR

TPRT

TTPTc

kkkkkk

s

kk

P

Psp

vv

v

Substituting,

KkJ/kg1.0451397.1

K)kJ/kg71.397(0.291)1( k

kRPR

kTkPTc p

(b) The cp is defined as cp = PT

h . Replacing the differentials by differences,

KkJ/kg1.045K390410

kJ/kg/28.011,34711,932K390410

K390410

kPa300

hKhThc

Pp

(Compare: Table A-2b at 400 K cp = 1.044 kJ/kg·K)

12-47

12-77 The temperature change of steam and the average Joule-Thompson coefficient during a throttlingprocess are to be estimated.Analysis The enthalpy of steam at 4.5 MPa and T = 300 C is h = 2944.2 kJ/kg. Now consider a throttlingprocess from this state to 2.5 MPa. The temperature of the steam at the end of this throttling process is

C72.273kJ/kg2.2944

MPa5.22T

hP

Thus the temperature drop during this throttling process is C26.2830072.27312 TTT

The average Joule-Thomson coefficient for this process is determined from

C/MPa13.14MPa4.52.5

C003273.72

kJ/kg7.3204hh PT

PT

12-78 The initial state and the final temperature of argon contained in a rigid tank are given. The mass of the argon in the tank, the final pressure, and the heat transfer are to be determined using the generalizedcharts.Analysis (a) The compressibility factor of argon at the initial state is determined from the generalized chart to be

Q

Ar-100 C1 MPa

18.0and95.0206.0

86.41

146.10.151

173

1

1

1

1

cr

1

cr

1

h

R

R

ZZ

PP

P

TT

T

Then,

kg135./kgm0.0342

m1.2

/kgm0342.0kPa1000

K)K)(173/kgmkPa081(0.95)(0.2

3

3

33

vV

vv

m

PZRTZRTP

(b) The specific volume of argon remains constant during this process, v2 = v 1. Thus,

kPa1531)4860)(315.0(

099.0

315.0

29.5kPa)K)(4860K)(151/kgmkPa(0.2081

/kgm0.0342/

808.10.151

273

cr2

2

3

3

crcr

2

cr

2

2

2

2

2

2

PPP

ZZ

P

PRT

TT

T

R

h

R

R

R

vv

(c) The energy balance relation for this closed system can be expressed as

)()()(

112212112212in

12in

systemoutin

TZTZRhhmPPhhmQuumUQ

EEE

vv

wherekJ/kg69.5710005203.0018.01512081.0ideal1212 21

hhZZRThh hhcr

Thus,kJ1251K)(0.95)(173)(0.99)(273K)kJ/kg(0.208157.69kg35.1inQ

12-48

12-79 Argon enters a turbine at a specified state and leaves at another specified state. Power output of theturbine and exergy destruction during this process are to be determined using the generalized charts.Properties The gas constant and critical properties of Argon are R = 0.2081 kJ/kg.K, Tcr = 151 K, and Pcr= 4.86 MPa (Table A-1).Analysis (a) The enthalpy and entropy departures of argon at the specified states are determined from thegeneralized charts to be

0044.1

86.47

97.3151600

11

1

1

cr

1

cr

1

sh

R

R

ZandZ

PP

P

TT

T

T0 = 25 CW·

P1 = 7 MPaT1 = 600 K V1 = 100 m/s

Ar = 5 kg/sm·

60 kW

Thus argon behaves as an ideal gas at turbine inlet. Also,

02.0and04.0206.0

86.41

85.1151280

22

2

2

cr

2

cr

2

sh

R

R

ZZ

PP

P

TT

T

P2 = 1 MPaT2 = 280 K V2 = 150 m/s Thus,

kJ/kg8.1676002805203.004.001512081.0ideal1212 21

hhZZRThh hhcr

The power output of the turbine is to be determined from the energy balance equation,

out

21

22

12out

outout2

222

11

outinsystemoutin

2)(

)2/()2/(

(steady)0

QVV

hhmW

WQVhmVhm

EEEEE

Substituting,

kW747.8kJ/s60/sm1000

kJ/kg12

m/s)(100m/s)(1508.167)kg/s5(

22

22

outW

(b) Under steady conditions, the rate form of the entropy balance for the turbine simplifies to

( ),

S S S S

ms ms QT

S S m s s QTb

in out gen system

out

outgen gen

out

0

1 2 2 20

0

0

The exergy destroyed during a process can be determined from an exergy balance or directly from its definition ,X Tdestroyed gen0S

0

out220gen0destroyed )(

TQ

ssmTSTX

where ideal1212 21ssZZRss ss

and KkJ/kg0084.071ln2081.0

600280ln5203.0lnln

1

2

1

2ideal12 P

PR

TT

css p

Thus, KkJ/kg0042.00084.0)02.0(0)2081.0(ideal1212 21ssZZRss ss

Substituting, kW366.K298

kW60KkJ/kg0.0042kg/s5K298destroyedX

12-49

12-80 EES Problem 12-79 is reconsidered. The problem is to be solved assuming steam is the workingfluid by using the generalized chart method and EES data for steam. The power output and the exergydestruction rate for these two calculation methods against the turbine exit pressure are to be plotted.Analysis The problem is solved using EES, and the results are tabulated and plotted below.

" Input Data "T[1]=600 [K] P[1]=7000 [kPa] Vel[1]=100 [m/s] T[2]=455 [K] P[2]=1000 [kPa] Vel[2]=150 [m/s] Q_dot_out=60 [kW] T_o=25+273 "[K]"m_dot=5 [kg/s] Name$='Steam_iapws' T_critical=647.3 [K] P_critical=22090 [kPa] Fluid$='H2O'

R_u=8.314M=molarmass(Fluid$)R=R_u/M

"****** IDEAL GAS SOLUTION ******""State 1"h_ideal[1]=enthalpy(Fluid$,T=T[1]) "Enthalpy of ideal gas"s_ideal[1]=entropy(Fluid$, T=T[1], P=P[1]) "Entropy of ideal gas""State 2"h_ideal[2]=enthalpy(Fluid$,T=T[2]) "Enthalpy of ideal gas"s_ideal[2]=entropy(Fluid$, T=T[2], P=P[2]) "Entropy of ideal gas"

"Conservation of Energy, Steady-flow: ""E_dot_in=E_dot_out"

m_dot*(h_ideal[1]+Vel[1]^2/2*convert(m^2/s^2,kJ/kg))=m_dot*(h_ideal[2]+Vel[2]^2/2*convert(m^2/s^2,kJ/kg))+Q_dot_out+W_dot_out_ideal

"Second Law analysis:""S_dot_in-S_dot_out+S_dot_gen = 0"m_dot*s_ideal[1] - m_dot*s_ideal[2] - Q_dot_out/T_o + S_dot_gen_ideal = 0

"Exergy Destroyed:"X_dot_destroyed_ideal = T_o*S_dot_gen_ideal

"***** COMPRESSABILITY CHART SOLUTION ******""State 1"Tr[1]=T[1]/T_criticalPr[1]=P[1]/P_criticalZ[1]=COMPRESS(Tr[1], Pr[1]) DELTAh[1]=ENTHDEP(Tr[1], Pr[1])*R*T_critical "Enthalpy departure"h_chart[1]=h_ideal[1]-DELTAh[1] "Enthalpy of real gas using charts"DELTAs[1]=ENTRDEP(Tr[1], Pr[1])*R "Entropy departure"s_chart[1]=s_ideal[1]-DELTAs[1] "Entropy of real gas using charts""State 2"

12-50

Tr[2]=T[2]/T_criticalPr[2]=P[2]/P_criticalZ[2]=COMPRESS(Tr[2], Pr[2]) DELTAh[2]=ENTHDEP(Tr[2], Pr[2])*R*T_critical "Enthalpy departure"DELTAs[2]=ENTRDEP(Tr[2], Pr[2])*R "Entropy departure"h_chart[2]=h_ideal[2]-DELTAh[2] "Enthalpy of real gas using charts"s_chart[2]=s_ideal[2]-DELTAs[2] "Entropy of real gas using charts"


m_dot*(h_chart[1]+Vel[1]^2/2*convert(m^2/s^2,kJ/kg))=m_dot*(h_chart[2]+Vel[2]^2/2*convert(m^2/s^2,kJ/kg))+Q_dot_out+W_dot_out_chart

"Second Law analysis:""S_dot_in-S_dot_out+S_dot_gen = 0"m_dot*s_chart[1] - m_dot*s_chart[2] - Q_dot_out/T_o + S_dot_gen_chart = 0

"Exergy Destroyed:"

X_dot_destroyed_chart = T_o*S_dot_gen_chart"[kW]"

"***** SOLUTION USING EES BUILT-IN PROPERTY DATA *****""At state 1"h_ees[1]=enthalpy(Name$,T=T[1],P=P[1])s_ees[1]=entropy(Name$,T=T[1],P=P[1])"At state 2"h_ees[2]=enthalpy(Name$,T=T[2],P=P[2])s_ees[2]=entropy(Name$,T=T[2],P=P[2])


m_dot*(h_ees[1]+Vel[1]^2/2*convert(m^2/s^2,kJ/kg))=m_dot*(h_ees[2]+Vel[2]^2/2*convert(m^2/s^2,kJ/kg))+Q_dot_out+W_dot_out_ees

"Second Law analysis:""S_dot_in-S_dot_out+S_dot_gen = 0"m_dot*s_ees[1] - m_dot*s_ees[2] - Q_dot_out/T_o + S_dot_gen_ees= 0

"Exergy Destroyed:"X_dot_destroyed_ees = T_o*S_dot_gen_ees

12-51

P2

[kPa]T2

[K]Woutchart

[kW]Woutees

[kW]Woutideal

[kW]Xdestroyedchart

[kW]Xdestroyedees

[kW]Xdestroyedideal

[kW]100 455 713.3 420.6 1336 2383 2519 2171200 455 725.2 448.1 1336 1901 2029 1694300 455 737.3 476.5 1336 1617 1736 1416400 455 749.5 505.8 1336 1415 1523 1218500 455 761.7 536.1 1336 1256 1354 1064600 455 774.1 567.5 1336 1126 1212 939700 455 786.5 600 1336 1014 1090 833800 455 799.1 633.9 1336 917.3 980.1 741.2900 455 811.8 669.3 1336 831 880.6 660.2

1000 455 824.5 706.6 1336 753.1 788.4 587.7

100 200 300 400 500 600 700 800 900 1000400

800

1200

1600

P[2] [kPa]

Wout;ees

[kW] Solution Method

EESEESChartChartIdeal gasIdeal gas

100 200 300 400 500 600 700 800 900 1000400

800

1200

1600

2000

2400

2800

P[2] [kPa]

Xdestroyed

[kW]

Solution MethodEESEESChartsChartsIdeal GasIdeal Gas

12-52

12-81E Argon gas enters a turbine at a specified state and leaves at another specified state. The poweroutput of the turbine and the exergy destruction associated with the process are to be determined using thegeneralized charts. Properties The gas constant and critical properties of argon are R = 0.04971 Btu/lbm.R, Tcr = 272 R, and Pcr = 705 psia (Table A-1E).Analysis (a) The enthalpy and entropy departures of argon at the specified states are determined from thegeneralized charts to be

0and0418.1

7051000

68.3272

1000

11

1

1

cr

1

cr

1

sh

R

R

ZZ

PP

P

TT

T

W·

P1 = 1000 psia T1 = 1000 R V1 = 300 ft/s

m·Ar

= 12 lbm/s

80 Btu/s

Thus argon behaves as an ideal gas at turbine inlet. Also,

02.0and04.0213.0

705150

P

838.1272500

22

2

2

cr

2

cr

2

sh

R

R

ZZP

P

TT

T

P2 = 150 psia T2 = 500 R V2 = 450 ft/s

Thus ,

Btu/lbm2.6310005001253.004.0027204971.0ideal12cr12 21

hhZZRThh hh

The power output of the turbine is to be determined from the energy balance equation,

out

21

22

12outoutout2

222

11

outinsystemoutin

2)()2/()2/(

(steady)0

QVV

hhmWWQVhmVhm

EEEEE

hp922Btu/s651.4

Btu/s80/sft25,037

Btu/lbm12

ft/s)(300ft/s)(4502.63)lbm/s21(

22

22

outW

(b) Under steady conditions, the rate form of the entropy balance for the turbine simplifies to

0

out22gengen

out,

out21

0systemgenoutin

)(0

0

TQ

ssmSSTQ

smsm

SSSS

b

The exergy destroyed during a process can be determined from an exergy balance or directly from its definition ,X Tdestroyed gen0S

0

out220gen0destroyed )(

TQ

ssmTSTX

where ideal1212 21ssZZRss ss

and RBtu/lbm00745.01000150ln04971.0

1000500ln1253.0lnln

1

2

1

2ideal12 P

PR

TT

css p

Thus RBtu/lbm00646.000745.0)02.0(0)04971.0(ideal1212 21ssZZRss ss

Substituting, sBtu5121 /.R535

Btu/s80RBtu/lbm0.00646lbm/s12R535destroyedX

12-53

12-82 An adiabatic storage tank that is initially evacuated is connected to a supply line that carries nitrogen. A valve is opened, and nitrogen flows into the tank. The final temperature in the tank is to bedetermined by treating nitrogen as an ideal gas and using the generalized charts, and the results are to be compared to the given actual value. Assumptions 1 Uniform flow conditions exist. 2 Kinetic and potential energies are negligible.Analysis We take the tank as the system, which is a control volume since mass crosses the boundary.Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h andinternal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance: m m m m m m miin out system out initial (since )2 0

Energy balance: E E E m h m ui iin out system 0 2 2

N2

V1 = 0.2 m3

Initiallyevacuated

10 MPa225 K

Combining the two balances: u2 = hi

(a) From the ideal gas property table of nitrogen, at 225 K we read

u h hi2 6 537@225 K kJ / kmol,

The temperature that corresponds to this u value is 2

(7.4% error)T2 314.8 K

(b) Using the generalized enthalpy departure chart, hi is determined to be

9.095.2

39.310

78.12.126

225ideal,

,

cr,

cr,

cru

iiih

iiR

iiR

TRhh

Z

PP

P

TT

T (Fig. A-29)

Thus,

and

kJ/kmol593,5

kJ/kmol593,52.126314.89.0537,69.0

2

crideal,

i

uii

hu

TRhh

Try T2 = 280 K. Then at PR2 = 2.95 and TR2 = 2.22 we read Z2 = 0.98 and cr2,2 /)( TRhh uideal = 0.55

Thus,

kJ/kmol283,5280314.898.0564,7

kJ/kmol564,72.126314.855.0141,855.0

222

crideal2,2

TZRhu

TRhh

u

u

Try T2 = 300 K. Then at PR2 = 2.95 and TR2 = 2.38 we read Z2 = 1.0 and cr2,2 /)( TRhh uideal = 0.50

Thus,

kJ/kmol704,5300314.80.1198,8

kJ/kmol198,82.126314.850.0723,850.0

222

crideal2,2

TZRhu

TRhh

u

u

By linear interpolation, (0.6% error)K294.72T

12-54

12-83 It is to be shown that dPdTdvv . Also, a relation is to be obtained for the ratio of specific

volumes v 2/ v 1 as a homogeneous system undergoes a process from state 1 to state 2. Analysis We take v = v (P, T). Its total differential is

dPP

dTT

dTP

vvv

Dividing by v,

dPP

dTT

d

TP

vv

vvv

v 11

Using the definitions of and ,

dPdTdvv

Taking and to be constants, integration from 1 to 2 yields

12121

2ln PPTTv

v

which is the desired relation.

12-84 It is to be shown that dPdTdvv . Also, a relation is to be obtained for the ratio of specific

volumes v 2/ v 1 as a homogeneous system undergoes an isobaric process from state 1 to state 2. Analysis We take v = v (P, T). Its total differential is

dPP

dTT

dTP

vvv

which, for a constant pressure process, reduces to

dTT

dP

vv

Dividing by v,

dTT

d

P

vvv

v 1

Using the definition of ,

dTdvv

Taking to be a constant, integration from 1 to 2 yields

or

121

2

121

2

exp

ln

TT

TT

v

v

v

v


12-55

12-85 The volume expansivity of water is given. The change in volume of water when it is heated atconstant pressure is to be determined.Properties The volume expansivity of water is given to be 0.207 10-6 K-1 at 20 C.Analysis We take v = v (P, T). Its total differential is

dPP

dTT

dTP

vvv


dTT

dP

vv

Dividing by v and using the definition of ,

dTdTT

d

P

vvv

v 1


or

121

2

121

2

exp

ln

TT

TT

v

v

v

v

Substituting the given values and noting that for a fixed mass V2/V1 = v2/v1,

3

1631212

m00000414.1

C1030K10207.0expm1exp TTVV

Therefore,3cm4.143

12 m00000414.0100000414.1= VVV

12-56

12-86 The volume expansivity of copper is given at two temperatures. The percent change in the volume of copper when it is heated at atmospheric pressure is to be determined.Properties The volume expansivity of copper is given to be 49.2 10-6 K-1 at 300 K, and be 54.2 10-6 K-1 at 500 K Analysis We take v = v (P, T). Its total differential is

dPP

dTT

dTP

vvv


dTT

dP

vv

Dividing by v and using the definition of ,

dTdTT

d

P

vvv

v 1


or

121

2

121

2

exp

ln

TT

TT

v

v

v

v

The average value of is 1666

21ave K107.512/102.54102.492/

Substituting the given values,

0104.1K300500K107.51expexp 1612

1

2 TTv

v

Therefore, the volume of copper block will increase by 1.04 percent.

12-57

12-87 It is to be shown that the position of the Joule-Thompson coefficient inversion curve on the T-Pplane is given by ( Z/ T)P = 0. Analysis The inversion curve is the locus of the points at which the Joule-Thompson coefficient is zero,

01v

v

Pp TT

c

which can also be written as

0P

ZRTT

TP

v (a)

since it is given that

PZRT

v (b)

Taking the derivative of (b) with respect to T holding P constant gives

ZT

TPR

TPZRT

T PPP

Z/v

Substituting in (a),

0

0

0

P

P

P

TZ

ZZTZT

PZRTZ

TZT

PTR


12-58

12-88 It is to be shown that for an isentropic expansion or compression process Pv k = constant. It is also tobe shown that the isentropic expansion exponent k reduces to the specific heat ratio cp/cv for an ideal gas. Analysis We note that ds = 0 for an isentropic process. Taking s = s(P, v), the total differential ds can be expressed as

0vvv

dsdPPsds

P (a)

We now substitute the Maxwell relations below into (a)

to get

0

and

vv

vv

v

dTPdP

T

TPs

TPs

ss

sPs

Rearranging,

00 vv

vv

dPdPdTPTdP

sss

Dividing by P, 01v

vdP

PPdP

s (b)

We now define isentropic expansion exponent k as

s

PP

kv

v

Substituting in (b), 0vvdk

PdP

Taking k to be a constant and integrating,

Thus,constant

constantlnconstantlnln

k

k

P

PkP

v

vv

To show that k = cp/cv for an ideal gas, we write the cyclic relations for the following two groups of variables:

)(11,,

)(11,,

dPT

sP

Tc

PT

sP

TsPTs

cTsT

cTsT

sTs

sT

p

sTP

sTsT vv

vv

v v

v

where we used the relations P

p TsTc

TsTc and

vv

Setting Eqs. (c) and (d) equal to each other,

sTsT

p TsT

cPT

sP

Tc

vvv

or,

sTsTsTsT

p PP

TTP

sPsT

sTP

Ps

cc

vv

vv

vv

v

butPP

PRTP TT

vv /

Substituting, kPPc

c

s

p

vv

v


12-59

12-89 EES The work done by the refrigerant 134a as it undergoes an isothermal process in a closed systemis to be determined using the tabular (EES) data and the generalized charts.

Analysis The solution using EES built-in property data is as follows:

kJ/kg.K2035.1kJ/kg35.280

MPa1.0C60

kJ/kg.K4828.0kJ/kg65.135

MPa3C60

2

2

2

2

1

1

1

1

su

PT

su

PT

kJ/kg95.40)65.13535.280(1.240)(

kJ/kg11.240)kJ/kg.K7207.0)(K15.27360(

kJ/kg.K0.72074828.02035.1

12EESEES

EES1EES

12EES

uuqw

sTq

sss

For the generalized chart solution we first determine the following factors using EES as

02281.0and03091.0,988.002464.0

059.41.0

8903.02.37415.333

383.4and475.4,1292.07391.0

059.43

8903.02.37415.333

2222

2

22

1111

1

11

sh

crR

crR

sh

crR

crR

ZZZ

PPP

TTT

ZZZ

PPP

TTT

Then,

kJ/kg.K3572.0kJ/kg.K)08148.0)(383.4(kJ/kg43.136K)74.2kJ/kg.K)(308148.0)(475.4(

11

cr11

RZsRTZh

s

h

kJ/kg.K001858.0kJ/kg.K)08148.0)(02281.0(kJ/kg94.0K)74.2kJ/kg.K)(308148.0)(03091.0(

22

cr22

RZsRTZh

s

h

KkgkJ2771.03

0.1kJ/kg.K)ln08148.0(ln1

2ideal /

PP

Rs

KkJ/kg0.6324)3572.0001858.0(2771.0)( 12idealchart ssss

kJ/kg70.210)kJ/kg.K6324.0)(K15.27360(chart1chart sTq

kJ/kg98.5317.11270.210

kJ/kg17.112)333)(08148.0)(1292.0()333)(08148.0)(988.0()43.13694.0(0)()(

chartchartchart

112212idealchart

uqw

RTZRTZhhhu

The copy of the EES solution of this problem is given next.

12-60

"Input data"T_critical=T_CRIT(R134a) "[K]"P_critical=P_CRIT(R134a) "[kpa]"T[1]=60+273.15"[K]"T[2]=T[1]"[K]"P[1]=3000"[kPa]"P[2]=100"[kPa]"R_u=8.314"[kJ/kmol-K]"M=molarmass(R134a)R=R_u/M"[kJ/kg-K]"

"***** SOLUTION USING EES BUILT-IN PROPERTY DATA *****"

"For the isothermal process, the heat transfer is T*(s[2] - s[1]):"DELTAs_EES=(entropy(R134a,T=T[2],P=P[2])-entropy(R134a,T=T[1],P=P[1]))q_EES=T[1]*DELTAs_EES

s_2=entropy(R134a,T=T[2],P=P[2])s_1=entropy(R134a,T=T[1],P=P[1])

"Conservation of energy for the closed system:"DELTAu_EES=intEnergy(R134a,T=T[2],p=P[2])-intEnergy(R134a,T=T[1],P=P[1])q_EES-w_EES=DELTAu_EESu_1=intEnergy(R134a,T=T[1],P=P[1])u_2=intEnergy(R134a,T=T[2],p=P[2])

"***** COMPRESSABILITY CHART SOLUTION ******""State 1"Tr[1]=T[1]/T_criticalpr[1]=p[1]/p_criticalZ[1]=COMPRESS(Tr[1], Pr[1]) DELTAh[1]=ENTHDEP(Tr[1], Pr[1])*R*T_critical"Enthalpy departure"Z_h1=ENTHDEP(Tr[1], Pr[1]) DELTAs[1]=ENTRDEP(Tr[1], Pr[1])*R "Entropy departure"Z_s1=ENTRDEP(Tr[1], Pr[1])

"State 2"Tr[2]=T[2]/T_criticalPr[2]=P[2]/P_criticalZ[2]=COMPRESS(Tr[2], Pr[2]) DELTAh[2]=ENTHDEP(Tr[2], Pr[2])*R*T_critical"Enthalpy departure"Z_h2=ENTHDEP(Tr[2], Pr[2]) DELTAs[2]=ENTRDEP(Tr[2], Pr[2])*R "Entropy departure"Z_s2=ENTRDEP(Tr[2], Pr[2])

"Entropy Change"DELTAs_ideal= -R*ln(P[2]/P[1]) DELTAs_chart=DELTAs_ideal-(DELTAs[2]-DELTAs[1])

"For the isothermal process, the heat transfer is T*(s[2] - s[1]):"q_chart=T[1]*DELTAs_chart

"Conservation of energy for the closed system:"DELTAh_ideal=0DELTAu_chart=DELTAh_ideal-(DELTAh[2]-DELTAh[1])-(Z[2]*R*T[2]-Z[1]*R*T[1])q_chart-w_chart=DELTAu_chart

12-61

SOLUTION

DELTAh[1]=136.43DELTAh[2]=0.94DELTAh_ideal=0DELTAs[1]=0.3572DELTAs[2]=0.001858DELTAs_chart=0.6324 [kJ/kg-K] DELTAs_EES=0.7207 [kJ/kg-K] DELTAs_ideal=0.2771 [kJ/kg-K] DELTAu_chart=112.17DELTAu_EES=144.7M=102 [kg/kmol]P[1]=3000 [kPa] P[2]=100 [kPa] pr[1]=0.7391Pr[2]=0.02464P_critical=4059 [kpa] q_chart=210.70 [kJ/kg] q_EES=240.11 [kJ/kg] R=0.08148 [kJ/kg-K

R_u=8.314 [kJ/kmol-K]s_1=0.4828 [kJ/kg-K] s_2=1.2035 [kJ/kg-K] T[1]=333.2 [K] T[2]=333.2 [K] Tr[1]=0.8903Tr[2]=0.8903T_critical=374.2 [K] u_1=135.65 [kJ/kg] u_2=280.35 [kJ/kg] w_chart=98.53 [kJ/kg] w_EES=95.42 [kJ/kg]Z[1]=0.1292Z[2]=0.988Z_h1=4.475Z_h2=0.03091Z_s1=4.383Z_s2=0.02281

12-62

12-90 The heat transfer, work, and entropy changes of methane during a process in a piston-cylinderdevice are to be determined assuming ideal gas behavior, using generalized charts, and real fluid (EES) data.Analysis The ideal gas solution: (Properties are obtained from EES)

State 1:

/kgm04834.0kPa4000

K15.273100)kJ/kg.K5182.0(

kJ/kg4685)15.273100)(5182.0()4492(kJ/kg.K22.10MPa4C,100

kJ/kg4492C100

3

1

11

111

111

11

PT

R

RThusPT

hT

v

State 2:

/kgm08073.0kPa4000

K15.273350)kJ/kg.K5182.0(

kJ/kg4093)15.273350)(5182.0()3770(kJ/kg.K68.11MPa4C,350

kJ/kg3770C350

3

2

22

222

222

22

PT

R

RThusPT

hT

v

kJ/kg129.56/kg0.04834)m-73kPa)(0.0804000()( 312ideal vvPw

kJ/kg721.70)4685()4093(56.129)( 12idealideal uuwq

kJ/kg1.4622.1068.1112ideal sss

For the generalized chart solution we first determine the following factors using EES as

2555.0and4318.0,9023.05413.0

39.74

227.12.304

373

111

cr

11

cr

11

sh

R

R

ZZZ

PP

P

TT

T

06446.0and1435.0,995.05413.0

39.74

048.22.304

623

222

cr

22

cr

22

sh

R

R

ZZZ

PP

P

TT

T

State 1:

kJ/kg4734)15.373)(5182.0)(9023.0()4560(kJ/kg456007.68)4492(

kJ/kg07.68K)04.2kJ/kg.K)(35182.0)(4318.0(

1111

1ideal1,1

cr11

RTZhuhhh

RTZh h

/kgm04362.04000

15.373)5182.0)(9023.0( 3

1

111 P

TRZv

kJ/kg.K09.101324.022.10kJ/kg.K1324.0kJ/kg.K)5182.0)(2555.0(

11,ideal1

11

sssRZs s

State 2:

kJ/kg4114)15.623)(5182.0)(995.0()3793(kJ/kg379362.22)3770(

kJ/kg62.22K)04.2kJ/kg.K)(35182.0)(1435.0(

2222

2ideal2,2

cr22

RTZhuhhh

RTZh h

12-63

/kgm08033.04000

15.623)5182.0)(995.0( 3

2

222 P

TRZv

kJ/kg.K65.1103341.068.11kJ/kg.K03341.0kJ/kg.K)5182.0)(06446.0(

22,ideal2

22

sssRZs s

Then,

kJ/kg146.84/kg0.04362)m-33kPa)(0.0804000()( 312chart vvPw

kJ/kg766.84)4734()4114(84.146)( 12chartchart uuwq

kJ/kg1.5609.1065.1112chart sss

The solution using EES built-in property data is as follows:

kJ/kg.K439.1kJ/kg82.39

/kgm04717.0

MPa4C100

1

1

31

1

1

su

PT

v

kJ/kg.K06329.0kJ/kg52.564

/kgm08141.0

MPa4C350

2

2

32

2

2

su

PT

v

kJ/kg136.96/kg0.04717)m-41kPa)(0.0814000()( 312EES vvPw

kJ/kg741.31)82.39(52.56497.136)( 12EESEES uuwq

kJ/kg1.50)439.1(06329.012EES sss

12-64

Fundamentals of Engineering (FE) Exam Problems

12-91 A substance whose Joule-Thomson coefficient is negative is throttled to a lower pressure. Duringthis process, (select the correct statement)(a) the temperature of the substance will increase.(b) the temperature of the substance will decrease.(c) the entropy of the substance will remain constant.(d) the entropy of the substance will decrease.(e) the enthalpy of the substance will decrease.

Answer (a) the temperature of the substance will increase.

12-92 Consider the liquid-vapor saturation curve of a pure substance on the P-T diagram. The magnitudeof the slope of the tangent line to this curve at a temperature T (in Kelvin) is(a) proportional to the enthalpy of vaporization hfg at that temperature,(b) proportional to the temperature T,(c) proportional to the square of the temperature T,(d) proportional to the volume change vfg at that temperature,(e) inversely proportional to the entropy change sfg at that temperature,

Answer (a) proportional to the enthalpy of vaporization hfg at that temperature,

12-93 Based on the generalized charts, the error involved in the enthalpy of CO2 at 350 K and 8 MPa if it isassumed to be an ideal gas is (a) 0 (b) 20% (c) 35% (d) 26% (e) 65%

Answer (d) 26%

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).

T=350 "K"P=8000 "kPa"Pcr=P_CRIT(CarbonDioxide)Tcr=T_CRIT(CarbonDioxide)Tr=T/TcrPr=P/PcrZ=COMPRESS(Tr, Pr) hR=ENTHDEP(Tr, Pr)

12-65

12-94 Based on data from the refrigerant-134a tables, the Joule-Thompson coefficient of refrigerant-134aat 0.8 MPa and 100 C is approximately(a) 0 (b) -5 C/MPa (c) 11 C/MPa (d) 8 C/MPa (e) 26 C/MPa

Answer (c) 11 C/MPa

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).

T1=100 "C"P1=800 "kPa"h1=ENTHALPY(R134a,T=T1,P=P1)Tlow=TEMPERATURE(R134a,h=h1,P=P1+100)Thigh=TEMPERATURE(R134a,h=h1,P=P1-100)JT=(Tlow-Thigh)/200

12-95 For a gas whose equation of state is P(v - b) = RT, the specific heat difference cp – cv is equal to(a) R (b) R – b (c) R + b (d) 0 (e) R(1 + v/b)

Answer (a) R

Solution The general relation for the specific heat difference cp - cv is

TPp

PT

Tccv

vv

2

For the given gas, P(v - b) = RT. Then,

bP

bRTP

bRTP

PR

Tb

PRT

T

P

vvvv

vv

2)(

Substituting,

RbP

TRb

PPRTcc p )(

22

vvv

12-96 ··· 12-98 Design and Essay Problems

thermosolutions chapter12

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